The slope of the tangent line passing through $A$ is $y' = 2x|_{x=1} = 2$. So the equation of the tangent line $AB$ is $y = 2x - 1$. Hence the coordinates of $B$ and $D$ are $B(0, -1)$, $D(\frac{1}{2}, 0)$. Thus $D$ is the midpoint of line segment $AB$.

Consider $P(x, y)$, $C(x_0, x_0^2)$, $E(x_1, y_1)$, $F(x_2, y_2)$. Then by $\frac{AE}{EC} = \lambda_1$, we know $x_1 = \frac{1+\lambda_1 x_0}{1+\lambda_1}$, $y_1 = \frac{1+\lambda_1 x_0^2}{1+\lambda_1}$. From $\frac{BF}{FC} = \lambda_2$, we get $x_2 = \frac{\lambda_2 x_0}{1+\lambda_2}$, $y_2 = \frac{-1+\lambda_2 x_0^2}{1+\lambda_2}$.

$$\frac{y - \frac{1 + \lambda_1 x_0^2}{1 + \lambda_1}}{\frac{-1 + \lambda_2 x_0^2}{1 + \lambda_2} - \frac{1 + \lambda_1 x_0^2}{1 + \lambda_1}} = \frac{x - \frac{1 + \lambda_1 x_0}{1 + \lambda_1}}{\frac{\lambda_2 x_0}{1 + \lambda_2} - \frac{1 + \lambda_1 x_0}{1 + \lambda_1}}$$

Simplifying it, we get
$$[(\lambda_2 - \lambda_1)x_0 - (1 + \lambda_2)]y = [(\lambda_2 - \lambda_1)x_0^2 - 3]x + 1 + x_0 - \lambda_2 x_0^2. \quad (1)$$

When $x_0 \neq \frac{1}{2}$, the equation of line $CD$ is
$$y = \frac{2x_0^2 x - x_0^2}{2x_0 - 1}. \quad (2)$$

From (1) and (2), we get
$$\begin{cases} x = \frac{x_0 + 1}{3}, \\ y = \frac{x_0}{3}. \end{cases}$$

Eliminating $x_0$, we get the equation of the trail of point $P$ as $y = \frac{1}{3}(3x-1)^2$.

When $x_0 = \frac{1}{2}$, the equation of $EF$ is $-\frac{3}{2}y = (\frac{1}{4}\lambda_2 - \frac{1}{4}\lambda_1 - 3)x + \frac{3}{2} - \frac{1}{4}\lambda_2$, the equation of $CD$ is $x = \frac{1}{2}$. Combining them, we conclude that $(x, y) = (\frac{1}{2}, \frac{1}{12})$ is on the trail of $P$. Since $C$ and $A$ cannot be congruent, $x_0 \neq 1, x \neq \frac{2}{3}$.

Therefore the equation of the trail is $y = \frac{1}{3}(3x-1)^2$, $x \neq \frac{2}{3}$.

From Solution I, the equation of $AB$ is $y = 2x - 1$, $B(0, -1)$, $D(\frac{1}{2}, 0)$. Thus $D$ is the midpoint of $AB$.

Set $\gamma = \frac{CD}{CP}$, $t_1 = \frac{CA}{CE} = 1+\lambda_1$, $t_2 = \frac{CB}{CF} = 1+\lambda_2$. Then $t_1+t_2=3$. Since $AD$ is a median of $\triangle ABC$, $S_{\triangle CAB} = 2S_{\triangle CAD} = 2S_{\triangle CBD}$ where $S_{\triangle}$ denotes the area of $\triangle$. But
$$\frac{1}{t_1 t_2} = \frac{CE \cdot CF}{CA \cdot CB} = \frac{S_{\triangle CEF}}{S_{\triangle CAB}} = \frac{S_{\triangle CEP}}{2S_{\triangle CAD}} + \frac{S_{\triangle CFP}}{2S_{\triangle CED}}$$
$$= \frac{1}{2} \left( \frac{1}{t_1 \gamma} + \frac{1}{t_2 \gamma} \right) = \frac{t_1 + t_2}{2t_1 t_2 \gamma} = \frac{3}{2t_1 t_2 \gamma},$$
so $\gamma = \frac{3}{2}$ and $P$ is the center of gravity for $\triangle ABC$.

Consider $P(x, y)$ and $C(x_0, x_0^2)$. Since $C$ is different from $A$, $x_0 \neq 1$. Thus the coordinates of the center of gravity $P$ are $x = \frac{0+1+x_0}{3} = \frac{1+x_0}{3}$, $x \neq \frac{2}{3}$, $y = \frac{-1+1+x_0^2}{3} = \frac{x_0^2}{3}$. Eliminating $x_0$, we get $y = \frac{1}{3}(3x-1)^2$. Thus the equation of the trail is $y = \frac{1}{3}(3x-1)^2$, $x \neq \frac{2}{3}$.